Chapter 10: One- and Two-Sample Tests of Hypotheses:Consider a population with some unknown parameter . We are interested in testing (confirming or denying) some conjectures about . For example, we might be interested in testing the conjecture that = o, where o is a given value.

10.1-10.3: Introduction +:·     A statistical hypothesis is a conjecture concerning (or a statement about) the population.·     For example, if is an unknown parameter of the population, we may be interested in testing the conjecture that > o for some specific value o.·     We usually test the null hypothesis: Ho: = o (Null Hypothesis)

Against one of the following alternative hypotheses:

H1:

  o > o < o  

 (Alternative Hypothesis or Research Hypothesis)

     Possible situations in testing a statistical hypothesis:

Ho is true Ho is false

Accepting Ho Correct Decision Type II error()

Rejecting Ho Type I error()

Correct Decision

Type I error = Rejecting Ho when Ho is trueType II error = Accepting Ho when Ho is falseP(Type I error) = P(Rejecting Ho | Ho is true) = P(Type II error) = P(Accepting Ho | Ho is false) =

·     The level of significance of the test: = P(Type I error) = P(Rejecting Ho | Ho is true) ·     One-sided alternative hypothesis:

Ho: = o or Ho: = o

H1: > o H1: < o

·     Two-sided alternative hypothesis:Ho: = o H1: o 

·     The test procedure for rejecting Ho (accepting H1) or accepting Ho (rejecting H1) involves the following steps:1.    Determining a test statistic (T.S.)2.    Determining the significance level 3.    Determining the rejection region (R.R.) and the acceptance region (A.R.) of Ho .R.R. of Ho depends on H1 and

·     H1 determines the direction of the R.R. of Ho ·     determines the size of the R.R. of Ho

H1: o

Two-sided alternative H1: > o

One-sided alternative H1: < o

One-sided alternative 4.    Decision:We reject Ho (accept H1) if the value of the T.S. falls in the R.R. of Ho , and vice versa.

10.5: Single Sample: Tests Concerning a Single Mean (Variance Known):Suppose that X1, X2, …, Xn is a random sample of size n from distribution with mean and (known) variance 2.Recall:

XXE )(

nXVar X

22)(

nNX ,~

nXZ

/

Test Procedure:

Hypotheses Ho: = o

H1: o

Ho: = o

H1: > o

Ho: = o

H1: < o

Test Statistic (T.S.)

N(0,1)~n/

XZ o

R.R.and A.R.

of Ho

Decision: Reject Ho (and accept H1) at the significance level if:

Z > Z /2 or Z < Z /2

Two-Sided Test

Z < Z

 One-Sided Test

Z > Z  One-Sided Test

Example 10.3:A random sample of 100 recorded deaths in the United States during the past year showed an average of 71.8 years. Assuming a population standard deviation of 8.9 year, does this seem to indicate that the mean life span today is greater than 70 years? Use a 0.05 level of significance.Solution:.n=100, =71.8, =8.9=average (mean) life spano=70Hypotheses:Ho: = 70H1: > 70T.S. :

X

02.2100/9.8708.71

/o

nX

Z

Level of significance:=0.05R.R.:

Z =Z0.05=1.645Z > Z =Z0.05=1.645Decision:Since Z=2.02 R.R., i.e., Z=2.02>Z0.05, we reject Ho at =0.05 and accept H1: > 70. Therefore, we conclude that the mean life span today is greater than 70 years. Example 10.4: A manufacturer of sports equipment has developed a new synthetic fishing line that he claims has a mean breaking strength of 8 kilograms with a standard deviation of 0.5 kilograms. Test the hypothesis that =8 kg against the alternative that 8 kg if a random sample of 50 lines is tested and found to have a mean breaking strength of 7.8 kg. Use a 0.01 level of significance.

Solution: .n=50, =7.8, =0.5,=0.01, /2 = 0.005= mean breaking strengtho=8Hypotheses:Ho: = 8H1: 8T.S. :

83.250/5.088.7

/o

nX

Z

X

Z /2=Z0.005=2.575 and Z /2= Z0.005= 2.575Decision:Since Z= 2.83 R.R., i.e., Z= 2.83 < Z0.005, we reject Ho at =0.01 and accept H1: 8. Therefore, we conclude that the claim is not correct.

10.7: Single Sample: Tests on a Single Mean (Variance Unknown):Suppose that X1, X2, …, Xn is a random sample of size n from normal distribution with mean and unknown variance 2.Recall:

1nv); t(~n/S

XT

n

ii nXXS

1

2 )1/()(

Test Procedure:

Hypotheses Ho: = o

H1: o

Ho: = o

H1: > o

Ho: = o

H1: < o

Test Statistic (T.S.)

R.R.and A.R.

of Ho

Decision: Reject Ho (and accept H1) at the significance level if:

T > t /2 or T < t /2

Two-Sided Test

T < t

 One-Sided Test

T > t  One-Sided Test

)1 t(n~n/S

XT o

Example 10.5:… If a random sample of 12 homes included in a planned study indicates that vacuum cleaners expend an average of 42 kilowatt-hours per year with a standard deviation of 11.9 kilowatt-hours, does this suggest at the 0.05 level of significance that the vacuum cleaners expend, on the average, less than 46 kilowatt-hours annually? Assume the population of kilowatt-hours to be normal.Solution:.n=12, =42, S=11.9, =0.05=average (mean) kilowatt-hours annual expense of a vacuum cleanero=46Hypotheses:Ho: = 46H1: < 46T.S. :

X

;16.112/9.11

4642n/S

XT o

=df= n1=11 t /2= t0.05= 1.796

Decision:Since T=1.16R.R. (T=1.16A.R.), we do not reject Ho at =0.05 (i.e, accept Ho: = 46) and reject H1: < 46. Therefore, we conclude that is not less than 46 kilowatt-hours.

10.8: Two Samples: Tests on Two Means:Recall: For two independent samples:

If and are known, then we have:21 2

2

N(0,1)~

nn

)()XX(Z

2

22

1

21

2121

If and are unknown but = =2, then we have:21 2

2 21 2

2

)2nnt(~

n1

n1S

)()XX(T 21

21p

2121

Where the pooled estimate of 2 is

2)1()1(

21

222

2112

nn

SnSnS p

The degrees of freedom of is =n1+n22.Now, suppose we need to test the null hypothesis Ho: 1 = 2 Ho: 1 2 = 0Generally, suppose we need to testHo: 1 2 = d (for some specific value d)Against one of the following alternative hypothesis

2pS

H1:1 2 d1 2 > d1 2 < d 

Hypotheses Ho: 1 2 = d

H1: 1 2 d

Ho: 1 2 = d

H1: 1 2 > d

Ho: 1 2 = d

H1: 1 2 < d

Test Statistic (T.S.)

N(0,1)~

nn

d)XX(Z

2

22

1

21

21

)2nnt(~

n1

n1S

d)XX(T 21

21p

21

{if = =2 is unknown} 21 2

2

{if and are known} 21 2

2

R.R.and A.R.

of Ho

Decision: Reject Ho (and accept H1) at the significance level if:

or or or

T.S. R.R.Two-Sided Test

T.S. R.R.One-Sided Test

T.S. R.R.One-Sided Test

Example10.6:An experiment was performed to compare the abrasive wear of two different laminated materials. Twelve pieces of material 1 were tested by exposing each piece to a machine measuring wear. Teen pieces of material 2 were similarly tested. In each case, the depth of wear was observed. The samples of material 1 gave an average wear of 85 units with a sample standard deviation of 4, while the samples of materials 2 gave an average wear of 81 and a sample standard deviation of 5. Can we conclude at the 0.05 level of significance that the mean abrasive wear of material 1 exceeds that of material 2 by more than 2 units? Assume populations to be approximately normal with equal variances.

Solution:Material 1 material 2 n1=12 n2=10 =85 =81 S1=4 S2=5Hypotheses: Ho: 1 = 2 + 2 (d=2) H1: 1 > 2 + 2Or equivalently, Ho: 1 2 = 2 (d=2) H1: 1 2 > 2Calculation:=0.05

1X 2X

05.2021012

)5)(110()4)(112(2

)1()1( 22

21

222

2112

nnSnSnS p

Sp=4.478= n1+n22=12+10 2 = 20t0.05 = 1.725T.S.:

04.1

101

121)478.4(

2)8185(11

)(

21

21

nnS

dXXT

p

Decision:Since T=1.04 A.R. (T=1.04< t0.05 = 1.725), we accept (do

not reject) Ho and reject H1: 1 2 > 2 at =0.05.

10.11 One Sample: Tests on a Single Proportion:

Recall:.p = Population proportion of elements of Type A in the

population

elementsof.noTotalAtypeofelementsof.no

BAA

n = sample size X = no. of elements of type A in the sample of size n. =Sample proportion elements of Type A in the sample p̂

nX

For large n, we have

npqnpX

npqppZ

ˆ

~ N(0,1) (Approximately, q=1p)

Hypotheses Ho: p = po

H1: p po

Ho: p = po

H1: p > po

Ho: p = po

H1: p < po

Test Statistic (T.S.)

R.R.and A.R.

of Ho

Decision: Reject Ho (and accept H1) at the significance level if:

Z > Z /2 or Z < Z /2

Two-Sided Test

Z < Z

 One-Sided Test

Z > Z  One-Sided Test

oo

o

oo

o

qnpnpX

nqppp

Z

ˆ

~N(0,1) (qo=1po)